A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant varepsilon_1$\varepsilon_1$ and varepsilon_2$\varepsilon_2$, as shown in figures. The distance between the plates is d$d$ and area of each plate is A$A$. If capacitance in first configuration and second configuration are C_1$C_1$ and C_2$C_2$ respectively, then fracC_1C_2$\frac{C_1}{C_2}$ is:
Illustrates two parallel plate configurations: stacked horizontally (series) and stacked vertically (parallel).Illustrates two parallel plate configurations: stacked horizontally (series) and stacked vertically (parallel).
Keywords:#dielectric parallel plate capacitor#capacitance series and parallel#JEE Main 2025 Morning Q8#harmonic arithmetic mean capacitance#dielectric capacitor#series parallel capacitor#dielectric constant ratio
More Electrostatics Previous-Year Questions — Page 5
Q212025Dielectrics and Capacitance
A parallel plate capacitor has charge 5times10^-6mathrm~C$5\times10^{-6}\mathrm{~C}$. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is 4times10^-6mathrm~C$4\times10^{-6}\mathrm{~C}$ then the dielectric constant of the slab is _______. [cite: 183, 184]
Numerical Answer.Answer: 5 to 5
Solution
### Related Formula
Q_textind = Qleft(1 - frac1Kright)$Q_{\text{ind}} = Q\left(1 - \frac{1}{K}\right)$ [cite: 813]
### Core Logic
Substitute the values given for free surface charge Q = 5 times 10^-6\ textC$Q = 5 \times 10^{-6}\ \text{C}$ and bound induced charge Q_textind = 4 times 10^-6\ textC$Q_{\text{ind}} = 4 \times 10^{-6}\ \text{C}$ into the equation: [cite: 183, 184, 814]
4 times 10^-6 = 5 times 10^-6 left(1 - frac1Kright)$4 \times 10^{-6} = 5 \times 10^{-6} \left(1 - \frac{1}{K}\right)$ [cite: 814]
frac45 = 1 - frac1K implies frac1K = 1 - frac45 = frac15$\frac{4}{5} = 1 - \frac{1}{K} \implies \frac{1}{K} = 1 - \frac{4}{5} = \frac{1}{5}$ [cite: 815]
K = 5$K = 5$ [cite: 815]
### Pattern Recognition
The fraction of charge induced on the dielectric face scales structurally as fracK-1K$\frac{K-1}{K}$[cite: 813, 815]. Observing a ratio of 4$4$ parts out of 5$5$ implies that the constant factor K$K$ must equal 5$5$ directly[cite: 815].
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q242025Electric Flux
The electric field in a region is given by vecmathrmE = (2hatmathrmi + 4hatmathrmj + 6hatmathrmk) times 10^3mathrmN / mathrmC$\vec{\mathrm{E}} = (2\hat{\mathrm{i}} + 4\hat{\mathrm{j}} + 6\hat{\mathrm{k}}) \times 10^{3}\mathrm{N} / mathrm{C}$ . The flux of the field through a rectangular surface parallel to x-z plane is 6.0mathrmNm^2mathrmC^-1$6.0\mathrm{Nm}^2\mathrm{C}^{-1}$ . The area of the surface is __________ mathrmcm^2$\mathrm{cm}^2$ . [cite: 195, 196]
Numerical Answer.Answer: 15 to 15
Solution
### Related Formula
phi = vecE cdot vecA$\phi = \vec{E} \cdot \vec{A}$ [cite: 827]
### Core Logic
A surface aligned parallel to the xtext-z$x\text{-}z$ plane possesses an area vector pointing completely orthogonal to it along the y$y$-axis direction, meaning vecA = Ahatj$\vec{A} = A\hat{j}$[cite: 196, 827]. Performing the dot product: [cite: 827]
phi = left[(2hati + 4hatj + 6hatk) times 10^3right] cdot (Ahatj) = 4 times 10^3 A$\phi = \left[(2\hat{i} + 4\hat{j} + 6\hat{k}) \times 10^3\right] \cdot (A\hat{j}) = 4 \times 10^3 A$ [cite: 195, 827]
Given that the net flux magnitude is 6.0\ textNm^2textC^-1$6.0\ \text{Nm}^2\text{C}^{-1}$ [cite: 196]:
6 = 4 times 10^3 A implies A = frac64 times 10^3 = 1.5 times 10^-3\ textm^2$6 = 4 \times 10^3 A \implies A = \frac{6}{4 \times 10^3} = 1.5 \times 10^{-3}\ \text{m}^2$ [cite: 828, 829]
Converting square meters to square centimeters (1\ textm^2 = 10^4\ textcm^2$1\ \text{m}^2 = 10^4\ \text{cm}^2$): [cite: 196, 830]
A = 1.5 times 10^-3 times 10^4 = 15\ textcm^2$A = 1.5 \times 10^{-3} \times 10^4 = 15\ \text{cm}^2$ [cite: 830]
### Pattern Recognition
Always focus exclusively on the specific field component matched to the surface orientation normal[cite: 827]. For an xtext-z$x\text{-}z$ plane match, only the hatj$\hat{j}$ coefficient creates flux[cite: 196, 827]. Do not miss the metric scale unit transition at the end (m^2 rightarrow cm^2$m^2 \rightarrow cm^2$)[cite: 196, 830].
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q182025Coulomb's Law
A small uncharged conducting sphere is placed in contact with an identical sphere but having 4 times 10^-8$4 \times 10^{-8}$ C charge and then removed to a distance such that the force of repulsion between them is 9 times 10^-3$9 \times 10^{-3}$ N. The distance between them is (Take frac14pivarepsilon_0$\frac{1}{4\pi\varepsilon_{0}}$ as 9 times 10^9$9 \times 10^{9}$ in SI units)
A. 2 cm
B. 3 cm
C. 4 cm
D. 1 cm
Solution
### Related Formula
F = frack q_1 q_2r^2$F = \frac{k q_1 q_2}{r^2}$
### Core Logic
When two identical conducting spheres are brought into contact, the total initial charge splits equally between them:
q_1 = q_2 = frac4 times 10^-8\ mathrmC + 02 = 2 times 10^-8\ mathrmC$q_1 = q_2 = \frac{4 \times 10^{-8}\ \mathrm{C} + 0}{2} = 2 \times 10^{-8}\ \mathrm{C}$
Given repulsion force, F = 9 times 10^-3\ mathrmN$F = 9 \times 10^{-3}\ \mathrm{N}$:
9 times 10^-3 = frac9 times 10^9 times (2 times 10^-8) times (2 times 10^-8)r^2$9 \times 10^{-3} = \frac{9 \times 10^{9} \times (2 \times 10^{-8}) \times (2 \times 10^{-8})}{r^2}$9 times 10^-3 = frac36 times 10^-7r^2 implies r^2 = frac36 times 10^-79 times 10^-3 = 4 times 10^-4$9 \times 10^{-3} = \frac{36 \times 10^{-7}}{r^2} \implies r^2 = \frac{36 \times 10^{-7}}{9 \times 10^{-3}} = 4 \times 10^{-4}$r = 2 times 10^-2\ mathrmm = 2\ mathrmcm$r = 2 \times 10^{-2}\ \mathrm{m} = 2\ \mathrm{cm}$Charge redistribution schematic for two spheres Q18
### Pattern Recognition
Identical spheres in contact distribute net charge equally due to symmetric capacitance sharing: q' = Q_texttotal / 2$q' = Q_{\text{total}} / 2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q12025Energy Stored in a Capacitor
Consider a parallel plate capacitor of area A (of each plate) and separation 'd' between the plates. If E is the electric field and epsilon_0$\epsilon_{0}$ is the permittivity of free space between the plates, then potential energy stored in the capacitor is :-
### Related Formula
The electrostatic energy density u$u$ stored in an electric field E$E$ is given by:
u = frac12epsilon_0E^2$u = \frac{1}{2}\epsilon_{0}E^{2}$
The total potential energy U$U$ stored in a volume V$V$ is:
U = u cdot V$U = u \cdot V$
### Core Logic
For a parallel plate capacitor, the volume between the plates where the electric field exists is the product of the plate area A$A$ and the plate separation d$d$:
V = Ad$V = Ad$
### Step 1: Calculating Stored Energy
Substitute the volume expression into the total energy equation:
U = left(frac12epsilon_0E^2
ight)(Ad)$U = \left(\frac{1}{2}\epsilon_{0}E^{2}
ight)(Ad)$U = frac12epsilon_0E^2Ad$U = \frac{1}{2}\epsilon_{0}E^{2}Ad$
### Pattern Recognition
Energy density times volume is a universal relation for field fields. Remember that volume is simply cross-sectional area multiplied by distance (Ad$Ad$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q182025Capacitance of a Parallel Plate Capacitor
A parallel plate capacitor was made with two rectangular plates, each with a length of l = 3$l = 3$ cm and breath of b = 1$b = 1$ cm. The distance between the plates is 3mu m$3\mu m$ Out of the following, which are the ways to increase the capacitance by a factor of 10?
A. l = 30$l = 30$ cm, b = 1 cm, d=1mu m$d=1\mu m$
B. l = 3$l = 3$ cm, b=1$b=1$ cm, d=30~mu m$d=30~\mu m$
C. l = 6$l = 6$ cm, b=5$b=5$ cm, d=3~mu m$d=3~\mu m$
D. l = 1$l = 1$ cm, b=1textcm$b=1\text{cm}$, d=10~mu m$d=10~\mu m$
E. l = 5text cm$l = 5\text{ cm}$, b=2$b=2$ cm, d=1mu m$d=1\mu m$
Choose the correct answer from the options given below :
A. C and E only
B. B and D only
C. A only
D. C only
Solution
### Related Formula
The capacitance of a parallel plate system is given by :
C = fracepsilon_0Ad = fracepsilon_0lbd$C = \frac{\epsilon_{0}A}{d} = \frac{\epsilon_{0}lb}{d}$
where l$l$ is length, b$b$ is breadth, and d$d$ is separation distance.
### Core Logic
Evaluate the initial capacitance base scaling parameter :
C_0 = fracepsilon_0 times 3text cm times 1text cm3mutextm = 1 times epsilon_0text units$C_{0} = \frac{\epsilon_{0} \times 3\text{ cm} \times 1\text{ cm}}{3\mu\text{m}} = 1 \times \epsilon_{0}\text{ units}$
We want to increase this initial baseline capacitance value by a factor of 10, meaning our target capacitance is 10epsilon_0$10\epsilon_{0}$.
### Step 1: Audit Options
Let's check the capacitance for options C and E :
* Option C: l=6text cm, b=5text cm, d=3mutextm$l=6\text{ cm}, b=5\text{ cm}, d=3\mu\text{m}$ .
C_C = fracepsilon_0 times 6 times 53 = 10epsilon_0text units (Correct)$C_{C} = \frac{\epsilon_{0} \times 6 \times 5}{3} = 10\epsilon_{0}\text{ units (Correct)}$
* Option E: l=5text cm, b=2text cm, d=1mutextm$l=5\text{ cm}, b=2\text{ cm}, d=1\mu\text{m}$ .
C_E = fracepsilon_0 times 5 times 21 = 10epsilon_0text units (Correct)$C_{E} = \frac{\epsilon_{0} \times 5 \times 2}{1} = 10\epsilon_{0}\text{ units (Correct)}$
### Pattern Recognition
Capacitance scales matching the geometric factor fracl cdot bd$\frac{l \cdot b}{d}$. Look for options where this ratio scales up to exactly 10 times the initial baseline value.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
More Electrostatics Questions — jee_main_2025_03_april_morning
We Map Every Repeating Question in Competitive Exams.
Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.
Select Your Target Exam
Choose an exam track below to find formulas per chapter and patterns.
Syncing Exam Intelligence
Mapping formulas and patterns across all tracks…
PATH A — FULL LENGTH PRACTICE
Full Mock Test Hub
Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.
Under Development
PATH B — TARGETED PRACTICE
Topic-wise Practice Hub
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
